The textbook says that
$$\int\frac{-mu}{\sqrt{1-\left(\frac{u}{c}\right)^{2}}}du$$
can be solved by inspection to give
$$mc^{2}\sqrt{1-\left(\frac{u}{c}\right)^{2}}$$
In simple steps, could anyone please explain how this is done.
Thank you
[Math] Integration by inspection – how
calculusintegration
Related Solutions
Hint: Complete the square: $$\int\sqrt{x-x^2}dx=\int\sqrt{-x^2+2\frac x2-\frac14+\frac14}dx=\int\sqrt{-\left(x-\frac 12\right)^2+\frac14}dx$$ Set $u=x-\frac 12$ to obtain $$\int\sqrt{-u^2+\frac14}du$$ I think you now how to deal with this one.
The inverse Laplace transform of $F(s)=\frac{s}{s^2+a^2}$ is given by
$$f(x)=\frac1{2\pi i}\int_{\alpha-i\infty}^{\alpha+i\infty}F(s)e^{sx}ds$$
where $\alpha=\Re (s)$.
The integral is carried out by contour integration.
The contour from $\alpha-i\infty$ to $\alpha+i\infty$ is referred to as the Bromwich contour, and $\alpha$ is taken to the right of all singularities in order to insure
$$\int_0^\infty e^{-\alpha x}\vert f(x)\vert dx<\infty$$ Hence, using the residue theorem and as $R\to\infty$ and so $T\to\infty$, we get $$f(x)=\sum_{j=1}^2 \operatorname{Res}\left(\frac{se^{sx}}{s^2+a^2},s_j\right),\quad s_1=ia,\quad s_2=-ia$$ so $$f(x)=\frac{ia e^{iax}}{2ia}+\frac{iae^{-iax}}{2ia}=cos(ax)\quad (x>0)$$
Best Answer
You can always check whether Wolfram|Alpha shows you the steps for an integration; it does in this case (click "Show Steps").
Note that the integrand is proportional to $f'(g(u))g'(u)$ with $f(x)=\sqrt x$ and $g(u)=1-\left(\frac uc\right)^2$. Generally, if an integrand is a product of some expression and the derivative of some part of that expression, try substituting for that part.
I wouldn't worry too much about the fact that it says "by inspection" and it doesn't feel like mere inspection to you. If you do enough substitutions, over time you'll start to "see" this immediately and will be able to write down the integral without going through the mechanics of the substitution.